Find the wavelength of radiation having frequency $4 \times 10^{11} \ kHz$.
- A$7.5 \times 10^{-4} \ m$
- B$7.5 \times 10^{-7} \ m$
- C$7.5 \times 10^{-5} \ m$
- D$7.5 \times 10^{-6} \ m$
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The ratio of the energy of a photon with wavelength $3000 \ \mathring{A}$ to that of a photon with wavelength $6000 \ \mathring{A}$ is:
Which emission transition series is obtained when an electron jumps from $n_2 = \infty$ to $n_1 = 1$?
How is an electron observed? Explain.
Easy
View Solution$A$ particular station of All India Radio,New Delhi,broadcasts on a frequency of $1,368 \, kHz$. The wavelength of the electromagnetic radiation emitted by the transmitter is : [speed of light $c = 3.0 \times 10^{8} \, m \, s^{-1}$]
Assertion : All photons possess the same amount of energy.
Reason : Energy of photon does not depend upon wavelength of light used.
Reason : Energy of photon does not depend upon wavelength of light used.
MediumAIIMS 1998
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